# Thread: Help with volume: solid of revolution

1. ## Help with volume: solid of revolution

The problem asks us to find the volume of the area bounded by y=-x^2+6x-8 and y=0 rotated around the x-axis by using the shells method.

I did this and got an answer of pi, to confirm the answer I used the disks method and got an answer of 16pi/15... please help by showing me how you do both methods.

2. Originally Posted by acfixerdude
The problem asks us to find the volume of the area bounded by y=-x^2+6x-8 and y=0 rotated around the x-axis by using the shells method.

I did this and got an answer of pi, to confirm the answer I used the disks method and got an answer of 16pi/15... please help by showing me how you do both methods.

You show me your work, and I'll show you mine.

3. ## ok

Originally Posted by cribby
You show me your work, and I'll show you mine.
the graph of the original equation rotated around the x axis kinda looks like a football... if I take shells to calculate volume, then I have shells centered around x=3 and from -y to y in height. So the height is 2y and the width is going to be 2pi*r where r=3-x if we're differentiating from 2 to 3. substitute for y and we get the integral from 2 to 3 of 2(-x^2+6x-8)*2pi(3-x)dx = pi

using disks, the area of one disk is going to be pi*r^2 where r is going to be equal to y. A=pi(-x^2+6x-8)^2.... I set up the integral from 2 to 4 of pi(-x^2+6x-8)^2 dx = 16pi/15...

What am I doing wrong?

4. Originally Posted by acfixerdude
the graph of the original equation rotated around the x axis kinda looks like a football... if I take shells to calculate volume, then I have shells centered around x=3 and from -y to y in height. So the height is 2y and the width is going to be 2pi*r where r=3-x if we're differentiating from 2 to 3. substitute for y and we get the integral from 2 to 3 of 2(-x^2+6x-8)*2pi(3-x)dx = pi

using disks, the area of one disk is going to be pi*r^2 where r is going to be equal to y. A=pi(-x^2+6x-8)^2.... I set up the integral from 2 to 4 of pi(-x^2+6x-8)^2 dx = 16pi/15...

What am I doing wrong?
I think any "wrongness" is in the algebra of your calculation. I did this by disks and had the same set-up you describe for this method, but got

$\int_2^4 \pi (-(x-3)^2+1)\,dx\,=\,\pi (-\frac{1}{3}(x-3)^3+x)|_2^4\,=\,\frac{4 \pi }{3}$:

5. Originally Posted by cribby
I think any "wrongness" is in the algebra of your calculation. I did this by disks and had the same set-up you describe for this method, but got

$\int_2^4 \pi (-(x-3)^2+1)\,dx\,=\,\pi (-\frac{1}{3}(x-3)^3+x)|_2^4\,=\,\frac{4 \pi }{3}$:
I don't understand how you set yours up, thanks for trying to help out... I spoke to my professor and he says that I was mistaken in trying to differentiate with respect to x and that to do it via shells I'd have to differentiate with respect to y... I can visualize it now and know why the shells don't work with x.

Thanks